To solve the problem, we need to analyze the system consisting of two persons and a boat. We will calculate the center of mass before and after the 55 kg person moves towards the 65 kg person.
### Step-by-Step Solution:
1. **Identify the System**:
- We have two persons with masses \( m_1 = 55 \, \text{kg} \) and \( m_2 = 65 \, \text{kg} \).
- The mass of the boat is \( m_b = 100 \, \text{kg} \).
- The length of the boat is \( L = 3.0 \, \text{m} \).
2. **Determine Initial Positions**:
- Let’s assume the left end of the boat is at position \( x = 0 \) m and the right end is at position \( x = 3.0 \) m.
- The 55 kg person is at the left end (position \( x_1 = 0 \) m).
- The 65 kg person is at the right end (position \( x_2 = 3.0 \) m).
- The center of mass of the boat is at its midpoint, which is \( x_b = 1.5 \) m.
3. **Calculate Initial Center of Mass**:
- The total mass of the system \( M = m_1 + m_2 + m_b = 55 + 65 + 100 = 220 \, \text{kg} \).
- The initial center of mass \( x_{cm, initial} \) can be calculated using the formula:
\[
x_{cm, initial} = \frac{m_1 x_1 + m_2 x_2 + m_b x_b}{M}
\]
Substituting in the values:
\[
x_{cm, initial} = \frac{(55 \times 0) + (65 \times 3) + (100 \times 1.5)}{220} = \frac{0 + 195 + 150}{220} = \frac{345}{220} \approx 1.568 \, \text{m}
\]
4. **Determine Final Positions**:
- When the 55 kg person walks to the 65 kg person, they both will be at the same position, which is at the right end of the boat (position \( x = 3.0 \) m).
- The center of mass of the boat remains at \( x_b = 1.5 \) m.
5. **Calculate Final Center of Mass**:
- The final center of mass \( x_{cm, final} \) can be calculated as:
\[
x_{cm, final} = \frac{m_1 x_f + m_2 x_f + m_b x_b}{M}
\]
where \( x_f = 3.0 \) m (the position where both persons are sitting).
\[
x_{cm, final} = \frac{(55 \times 3) + (65 \times 3) + (100 \times 1.5)}{220} = \frac{165 + 195 + 150}{220} = \frac{510}{220} \approx 2.318 \, \text{m}
\]
6. **Calculate the Shift in Center of Mass**:
- The shift in the center of mass \( \Delta x_{cm} \) is given by:
\[
\Delta x_{cm} = x_{cm, final} - x_{cm, initial} = 2.318 - 1.568 \approx 0.75 \, \text{m}
\]
### Conclusion:
The center of mass of the system shifts by approximately \( 0.75 \, \text{m} \).
